Ask Question
13 December, 15:46

Give the percent yield when 162.8 g of CO2 are formed from the reaction of excess amount of

C8H18 and with 218.0 grams of O2

+2
Answers (2)
  1. 13 December, 16:05
    0
    84.86%

    Explanation:

    Step 1:

    We'll begin by writing a balanced equation for the reaction between C8H18 and O2 to produce CO2. This is illustrated below:

    2C8H18 + 25O2 - > 16CO2 + 18H2O

    Step 2:

    Now, let us calculate the mass of O2 that reacted and the mass of CO2 produced from the balanced equation above. This is illustrated below:

    Molar Mass of O2 = 16x2 = 32g/mol

    Mass of O2 from the balanced equation = 25 x 32 = 800g

    Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

    Mass of CO2 from the balanced equation = 16 x 44 = 704g

    Therefore the mass of O2 that reacted from the balanced equation is 800g

    The mass of CO2 produced from the balanced equation is 704g

    Step 3:

    Determination of the theoretical yield of CO2. This is illustrated below:

    From the balanced equation above,

    800g of O2 reacted to produced 704g of CO2.

    Therefore, 218g of O2 will react to produce = (218 x 704) / 800 = 191.84g of CO2.

    Therefore, the theoretical yield of CO2 is 191.84g

    Step 4:

    Determination of the percentage yield of CO2. This is illustrated below:

    Actual yield = 162.8g

    Theoretical yield = 191.84g

    Percentage yield = ?

    Percentage yield = Actual yieldm/Theoretical yield x100

    Percentage yield = 162.8/191.84 x100

    Percentage yield = 84.86%

    Therefore, the percentage yield of CO2 is 84.86%
  2. 13 December, 19:33
    0
    The percent yield of the reactions is 84.8 %

    Explanation:

    Step 1: data given

    Mass of CO2 formed = 162.8 grams

    Mass of O2 = 218.0 grams

    C8H18 is in excess

    Molar mass O2 = 32.0 g/mol

    Molar mass of CO2 = 44.01 g/mol

    Step 2: The balanced equation

    2C8H18 + 25O2 → 16CO2 + 18H2O

    Step 3: Calculate moles O2

    Moles O2 = 218.0 grams / 32.0 g/mol

    Moles O2 = 6.8125 moles

    Step 4: Calculate moles CO2

    For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

    For 6.8125 moles O2 we'll have 16/25 * 6.8125 = 4.36 moles CO2

    Step 5: Calculate mass CO2

    Mass CO2 = moles CO2 * molar mass CO2

    Mass CO2 = 4.36 moles ¨¨ 44.01 g/mol

    Mass CO2 = 191.9 grams

    Step 6: Calculate the percent yield

    Percent yield = (actual mass / theoretical mass) * 100 %

    Percent yield = (162.8 grams / 191.9 grams) * 100 %

    Percent yield = 84.8 %

    The percent yield of the reactions is 84.8 %
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Give the percent yield when 162.8 g of CO2 are formed from the reaction of excess amount of C8H18 and with 218.0 grams of O2 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers